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I have a problem with the following exercise.

Let $(\Omega, \mathcal{H}, \mathbb{P})$ be the probability space defined by $\Omega = (0,1)$, $\mathcal{H} = \mathcal{B}(0,1)$ and $\mathbb{P}$ is the Lebesgue measure. Consider the following random variables: $X(\omega)=\omega^{\,2}$, $Y(\omega)=\omega(1-\omega)$ for each $\omega \in \Omega$. I have to calculate $\mathbb{E}[X|Y]$ but I do not know how to proceed.

Every suggestion or help is appreciated.

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  • $\begingroup$ just a hunch, not 100% sure -- if you know $y = w(1-w)$ you can find $w=w(y)$ and from it compute $w^2$? $\endgroup$ – gt6989b Sep 19 '17 at 17:52
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For every $B\in\mathcal B$ we have the equality:$$\int^{0.5}_{-0.5}z1_B\left(0.25-z^2\right)dz=0\tag1$$

Substituting $\omega=0.5-z$ leads to:$$\int^1_0\left(\omega-0.5\right)1_B\left(\omega\left(1-\omega\right)\right)d\omega=0\tag2$$Of course we have: $$\omega-0.5=\omega^2-\left[0.5-\omega(1-\omega)\right]=X(\omega)-\left[0.5-Y(\omega)\right]\tag3$$ so $(2)$ is equivalent with:$$\int^1_0X\left(\omega\right)1_B\left(Y\left(\omega\right)\right)d\omega=\int^1_0\left[0.5-Y(\omega)\right]1_B(Y(\omega))d\omega\tag4$$

We can write this also as:$$\int_{\{Y\in B\}}X(\omega)d\omega=\int_{\{Y\in B\}}0.5-Y(\omega)d\omega\tag5$$

This being true for every $B\in\mathcal H$ states exactly that: $$\mathbb E(X\mid Y)=0.5-Y$$

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  • $\begingroup$ Thank you for the reply. Could you explain the steps from "or equivalently:" onwards? $\endgroup$ – LJG Sep 19 '17 at 21:31
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    $\begingroup$ Which of the steps $1\to2$,$2\to3$,$3\to4$, $4\to5$ (or something in between) needs further explaining? $\endgroup$ – drhab Sep 20 '17 at 8:18
  • $\begingroup$ I reread it and now I think I understand why you replace $X(\omega)$ with $0.5-Y(\omega)$ in (4). I guess (1) comes from your idea/observation. How did you write it? $\endgroup$ – LJG Sep 20 '17 at 22:05
  • $\begingroup$ I started with finding an expression for $X$ on base of $Y=y$. Actually there are two expressions and taking their average gives $X=0.5-y$. That was a hint. It is hard to describe how I arrived at $(1)$ and I can imagine that in my answer it seems as if it is some gift from heaven. For a big deal it is a matter of intuiton and mathematical maturity. I cannot hand over a recipe. Sorry for that. $\endgroup$ – drhab Sep 21 '17 at 7:48

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